Answer:
With garlic treatment, the mean change in LDL cholesterol is not greater than 0.
Step-by-step explanation:
The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.
In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering cholesterol.
A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.
The hypothesis for the test can be defined as follows:
<em>H₀</em>: With garlic treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.
<em>Hₐ</em>: With garlic treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.
The information provided is:
![\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01](https://tex.z-dn.net/?f=%5Cbar%20d%3D0.40%5C%5CSD_%7Bd%7D%3D16.2%5C%5C%5Calpha%20%3D0.01)
Compute the test statistic value as follows:
![t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20d%7D%7BSD_%7Bd%7D%2F%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B0.40%7D%7B16.2%2F%5Csqrt%7B81%7D%7D%5C%5C%5C%5C%3D0.22)
The test statistic value is 0.22.
Decision rule:
If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.
Compute the <em>p</em>-value of the test as follows:
![p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132](https://tex.z-dn.net/?f=p-value%3DP%28t_%7Bn-1%7D%3E0.22%29%5C%5C%3DP%28t_%7B80%7D%3E0.22%29%5C%5C%3D0.4132)
*Use a <em>t</em>-table.
The <em>p</em>-value of the test is 0.4132.
<em>p-</em>value= 0.4132 > <em>α</em> = 0.01
The null hypothesis was failed to be rejected.
Thus, it can be concluded that with garlic treatment, the mean change in LDL cholesterol is not greater than 0.